# Scalar coupled to Gauss-Bonnet invariant vs Horndeski theory

So here it is a somewhat tormenting question. The first statement will be a little specific but then I will make clear what the jargon indicates.

How can we show that a Lagrangian made of a scalar field linearly coupled to the Gauss-Bonnet invariant $$\mathcal{R}^2_{GB}\,=\,R^{abcd}R_{abcd}-4R^{ab}R_{ab}+R^2$$ is equivalent to the Horndeski Lagrangian where the function $$G_5(X)$$ is chosen to be $$\log|X|$$ and the other $$G_i$$ are zero?

Explicitly we have to prove that: \begin{align*} \sqrt{-g}\phi\mathcal{R}^2_{GB}=\sqrt{-g}\left[ G_5(X)G_{ab}\nabla^a\nabla^b\phi-\frac{G_{5X}(X)}{6}\Big((\nabla_a\nabla^a\phi)^3\\-3\;(\nabla_c\nabla^c\phi)\;\nabla_a\nabla^b\phi\;\nabla_b\nabla^a\phi+2\;\nabla_a\nabla^b\phi\;\nabla_b\nabla^c\phi\;\nabla_c\nabla^a\phi\Big)\right] \end{align*} when $$G_5(X)=\log|X|$$ ,

where $$G_{ab}$$ is the Einstein tensor, $$X=-\frac{\nabla_a\phi\nabla^a\phi}{2}$$ and the subscript $$X$$ indicates derivation with respect to $$X$$ .

This equivalence was presented in this paper (appendix A, page 14), but no proof is to be found in the literature to my knowledge. The authors confirmed in more occasions that it holds in general and I guess that it holds up to total derivatives.

I verified it in a spherically symmetric static case by looking at the explicit expressions, but for now I was not able to show it in general.

Maybe somebody did it?